为什么使用`dyn Trait`的类型别名时出现尺寸错误?

问题描述 投票:-1回答:1

此函数编译:

fn edit<S: AsRef<str>>(w: S) {}

如果我键入通用参数别名:

type Word = dyn AsRef<str>;

fn edit(w: Word) {}

我收到一个错误:

error[E0277]: the size for values of type `(dyn std::convert::AsRef<str> + 'static)` cannot be known at compilation time
 --> src/lib.rs:3:9
  |
3 | fn edit(w: Word) {}
  |         ^ doesn't have a size known at compile-time
  |
  = help: the trait `std::marker::Sized` is not implemented for `(dyn std::convert::AsRef<str> + 'static)`
  = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
  = note: all local variables must have a statically known size
  = help: unsized locals are gated as an unstable feature

为什么会这样?

rust size type-alias
1个回答
1
投票

所有函数在编译时都需要知道其参数的大小。但是,您正在使用w,其大小无法在编译时确定。为了能够执行动态调度,您需要使用特征对象。您可以通过两种方式完成此操作。

© www.soinside.com 2019 - 2024. All rights reserved.